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Let's look at the datasheet for an MMBT3904, just for example. The absolute maximum section talks mostly about maximum voltage differences, and a single current limit - the collector current.

I'm used to using these, and similar BJTs as saturated switches. And I get that once you have a base current that is sufficient that the Hfe causes the collector current to max out... I'm good to go.

When I was a kid, I didn't know this stuff as well as I do now (which says a lot), so I tried setting up saturated BJT switches with no base resistor. So I'd pump way, way too much current from the base to emitter and blow it apart. I'm older and wiser now.

But the question still stands: From the absolute maximums on the datasheet, how do I calculate what is effectively the absolute minimum base resistor? That is, the absolute maximum amount of base current before the blue smoke is released?

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  • \$\begingroup\$ There's no point in saturating the transistor beyond full saturation. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Mar 6, 2015 at 18:57
  • \$\begingroup\$ @IgnacioVazquez-Abrams Clearly. The question is largely an academic one. \$\endgroup\$
    – nsayer
    Commented Mar 6, 2015 at 19:07
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    \$\begingroup\$ @IgnacioVazquez-Abrams , nsayer: As a BJT transistor is driven with decreasing forced Beta (ie higher base drive per collector current) Vsat falls and values much lower than normally achieved can be obtained. I have used a transistor with Ibase ~= 10 x Icollector!!! to obtain an extremely low Vsat - in the mV range. This was used to turn on a PNP transistor feeding a resistive divider for an ADC circuit, thereby ensuring very low error from the 'switch'. As ADC divider current was a small fraction of a mA and operation was only occasional the mA or so of drive current was entirely acceptable. \$\endgroup\$
    – Russell McMahon ♦
    Commented Mar 6, 2015 at 21:22
  • \$\begingroup\$ If I was doing the same task nowadays I'd use a suitably low Rdson MOSFET. \$\endgroup\$
    – Russell McMahon ♦
    Commented Mar 6, 2015 at 21:24

2 Answers 2

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I suspect that power dissipation would be the limiting factor.

Normally you'd only be concerned with (Vce x Ic) since (Vbe x Ib) would be insignificant in comparison.
But if you're driving the hell out of Ib it'll become far more of a significant term.

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  • \$\begingroup\$ So in my youth, by not limiting the base current with a resistor, the transistors blew simply because the effectively unlimited current exceeded the power dissipation maximum of the device? That would make sense. \$\endgroup\$
    – nsayer
    Commented Mar 6, 2015 at 19:15
  • \$\begingroup\$ When you work out that number it far exceeds the max collector current though, which intuitively just seems wrong. \$\endgroup\$
    – Trevor_G
    Commented Feb 1, 2018 at 21:37
  • \$\begingroup\$ @Trevor_G - At that point you've gone well into saturation, so the Ic/Ib Hfe gain doesn't really mean anything anymore and Ic will probably be determined more by the surrounding circuitry and power supply. \$\endgroup\$
    – brhans
    Commented Feb 1, 2018 at 21:44
  • \$\begingroup\$ Yes I know, but Ic max has nothing to do with Hfe either. It's when the bonding wire melts. Calculating Ib max from the power dissipation at .7v gets you higher current than Ic max. \$\endgroup\$
    – Trevor_G
    Commented Feb 1, 2018 at 21:47
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The 2N3904, presumably the same die, lists maximum peak base current as 100mA.

Since that presumption is not necessarily correct, I would suggest limiting peak current to a fraction of that (maybe 20-50mA) and continuous base current to maybe 20mA (1/10 of the maximum collector current).

Remember to take the power dissipation due to the base current into account- Vbe can be in excess of 1V at higher currents, so it will be significant in an SOT-23 package.

Looking at similar parts that I tend to use, the MMBT4401 is specified at 50mA base current and has 1.2V Vbe(max). The MMBT3904 will have a higher Vbe drop. The 8050 has a 1.2V Vbe(max) at 80mA base current (again the MMBT3904 will have much higher Vbe drop).

There's generally no point in having more base current than about Ic/10, so I would say that determines the minimum practical base resistor. If you're switching a 100mA relay coil using a 5V drive then the minimum practical resistor is about 400 ohms, which is well below the current that would likely cause long term damage, let alone immediate damage.

Note that the absolute maximum is not a target you should aim for, it's a warning of something specified by the manufacturer that you should stay well away from if you want reliability and should look carefully at the measurement conditions in case they are optimistic (for example, if abs max of something is specified at 25°C and your application sees 80°C you may have to derate the number considerably to be safe).

If there is no abs max of something specified that doesn't mean you have carte blanche to go as high as you want, it simply means the manufacturer didn't think it was important specify it, possibly because in most applications the engineers don't need to worry about it (as in this case).

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  • \$\begingroup\$ I never imagined it was ok to pump excess current through BE junction. I just want to learn how I can insure that my current designs aren't stressing things excessively. It's nice to get confirmation in the form of abs. max. ratings that are way far away from where you are. \$\endgroup\$
    – nsayer
    Commented Mar 6, 2015 at 20:03
  • \$\begingroup\$ @nsayer: See "forced beta" electronics.stackexchange.com/questions/311243/… which exactly answers this question. \$\endgroup\$
    – bkausbk
    Commented Dec 16, 2020 at 15:05

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